Question

A 135-g sample of a metal requires 2.50kj to change its temperature from 19.5 degrees Celsius to 100.0 degrees Celsius. What is the specific heat of this metal?

Answer 1

Q = m x c x ΔT

2500 = 0.135 x C x 80.5

2500 = 10.8765 x C

C = 230.043 J/Kg.K

hope this helps

Answer 2

Answer: The specific heat of the metal is $0.23J/g^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed=$2.50kJ=2500J$

m= mass of substance = 135 g

c = specific heat capacity = ?

Initial temperature = $T_i$ = 19.5°C

Final temperature = $T_f$  =100.0°C

Change in temperature ,$\Delta T=T_f-T_i=(100-19.5)^0C=80.5^0C$

Putting in the values, we get:

$2500=135\times c\times 80.5^0C$

$c=0.23J/g^0C$

The specific heat of the metal is $0.23J/g^0C$